Identify the compound that will react with Hinsberg's reagent to give a solid which dissolves in alkali.

Solution & Explanation

\textbf{Answer:} (A) Only a primary amine forms an $N$-substituted sulfonamide bearing an acidic $\mathrm{N}-\mathrm{H}$, which dissolves in alkali. \textbf{Solution:} Hinsberg's reagent is benzenesulfonyl chloride, $\mathrm{C}{}_6\mathrm{H}{}_5\mathrm{SO}{}_2\mathrm{Cl}$. With a primary amine it forms an $N$-alkylbenzenesulfonamide: $(\mathrm{CH}{}_3){}_2\mathrm{CH}-\mathrm{NH}{}_2+\mathrm{C}{}_6\mathrm{H}{}_5\mathrm{SO}{}_2\mathrm{Cl}\to \mathrm{C}{}_6\mathrm{H}{}_5\mathrm{SO}{}_2-\mathrm{NH}-\mathrm{CH}(\mathrm{CH}{}_3){}_2$. The remaining $\mathrm{N}-\mathrm{H}$ is acidic (the $-\mathrm{SO}{}_2-$ group withdraws electron density), so this solid dissolves in alkali ($\mathrm{KOH}$) to give a soluble salt. A secondary amine would give a sulfonamide with no $\mathrm{N}-\mathrm{H}$, hence a solid insoluble in alkali. Tertiary amines (B, D) have no $\mathrm{N}-\mathrm{H}$ to react and give no sulfonamide. The nitro compound (C) is not an amine and does not respond. Therefore the primary amine isopropylamine (A) is correct.